Question

30. Solve the following quadratic equation: 2x2 + 4x –6 = 0 Using the quadratic formula:

Answer 1

2x2 + 4x = 6

is that 2x² + 4x = 6?

if yes,

2x² + 4x = 6
send the 6 to the left:
2x² + 4x - 6 = 0
divide everything by 2 to simplify things (here all the numbers are factors of 2 so we can safely divide them):

x² + 2x - 3 = 0, now solve the quadratic equation:
u should look for 2 numbers which when multiplied together give u -3 and when added give u +2-->

(x +3 ) ( x- 1) = 0

So, either x +3 = 0 or x - 1 =0
either x = -3 or x = 1.

I hope this helps :))

Answer 2

this might be a little bit easier

Answer 3

thanks

Answer 4

and yes that is it ur awesome thank you :D

Answer 5

oh could you help me with two more problems?

Answer 6

First of all, if the problem specifically says that you are to solve using the QF, then you should be using the QF, not factoring. Sometimes (not always), either method will work. But if the problem says to use a particular method, then if you were in my class, you would lose points for NOT following that direction.

More importantly, the solution above is WRONG.
\[2x^2 + 4x –6 \neq(x +3 ) ( x- 1)\]
Like I always try to pound into my students' heads: CHECK FACTORING WITH MULTIPLICATION!!

\[(x +3 ) ( x- 1)=x^2+2x-3\]

Answer 7

i tried to make it easy so that he can understand

Answer 8

Sorry, I see now that you divided through by 2, I mis-read the solution.... the factoring was not wrong.

But still, QF isn't that hard - know what a, b and c are and then plug & chug. :) If the problem says to use QF, then that's what he should be using. :)

Answer 9

27. Show all work for finding the solution to this system of equations using substitution.
y = 2x + 4
y = 4x – 2

could yal help me with this also?

Answer 10

Both equations are y=..... so set the right-hand-sides equal to each other and solve for x. Then plug that value for x back into either equation to get y.